On time evolution of density operator (matrix) in quantum mechanics Suppose I have at time $t=0$ a statistical ensemble of quantum states
$$\{|\psi_n\rangle\langle\psi_n|\}_{n=1}^N$$
The probability of finding the result $a$ for an observable $A$ in this ensemble is given by the Born rule (assuming discrete spectrum, without degeneracy):
$$ \sum_{n=1}^N p_n \langle \psi_n | \Pi_a|\psi_n \rangle = \operatorname{Tr}\left[\rho\,\Pi_a\right]$$
where $\Pi_a$ is the projector in the $a$-eigenspace of $A$, and
$$
\rho:= \sum_{n=1}^N p_n |\psi_n\rangle\langle\psi_n|
$$
is the density operator or density matrix.
At time $t>0$, this density $\rho$ takes the form
$$
\rho(t):= \sum_{n=1}^N p_n |\psi_n(t)\rangle\langle\psi_n(t)|
$$
what puzzles me the most is that those $p_n$ stay the same. Why should I expect that? Why couldn't the probabilities change as each state $|\psi_n(t)\rangle\langle\psi_n(t)|$ evolves with time?
 A: The idea is really just that you start out with a statistical ensemble, i.e. the quantum system is in the state $\lvert \psi_n\rangle$ with probability $p_n$ at the start. The density matrix $\sum_n p_n\lvert \psi_n\rangle\langle \psi_n\rvert$ is just a way to summarize this information into a single object. The notion of density matrices does not add any new "mechanics" to quantum mechanics.
Each of these states evolves in time like quantum states normally do, i.e. $\lvert \psi_n(t)\rangle = U(t)\lvert \psi_n\rangle$ with $U(t)$ the ordinary time evolution operator. So if the system started in the state $\lvert \psi_n(0)\rangle$, of course it is in the state $\lvert \psi_n(t)\rangle$ at time $t$. Since it had probability $p_n$ to start in the state $\lvert \psi_n(0)\rangle$, it still has probability $p_n$ to be in $\lvert \psi_n(t)\rangle$ at time $t$, and the density matrix at time $t$ is $\sum_n p_n\lvert \psi_n(t)\rangle\langle \psi_n(t)\rvert$. There isn't anything more to this, the probabilities here just act like classical probabilities.
